2012 fermat contest - · pdf file2012 fermat contest solutions page 2 1. since 60 8 = 60 8 =...

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2012 Fermat Contest(Grade 11)

Thursday, February 23, 2012(in North America and South America)

Friday, February 24, 2012(outside of North America and South America)

Solutions

©2011 University of Waterloo

2012 Fermat Contest Solutions

1. Since 608

= 60÷ 8 = 7.5, then this choice is not equal to a whole number.Note as well that 60

12= 5, 60

5= 12, 60

4= 15, and 60

3= 20 are all whole numbers.

Answer: (B)

2. Simplifying the left side of the equation, we obtain 5 = 6− x.Therefore, x = 6− 5 = 1.

Answer: (C)

3. Since JFG is a straight line, then ∠HFG = 180◦ − ∠HFJ = 180◦ − 110◦ = 70◦.Since 4FGH is isosceles with HF = HG, then ∠HGF = ∠HFG = 70◦.Since the sum of the angles in 4FGH is 180◦, then 70◦ +70◦ +x◦ = 180◦, and so 140+x = 180or x = 40.

Answer: (E)

4. Simplifying inside the brackets first, (1 + 13)(1 + 1

4) = (4

3)(5

4) = 20

12= 5

3.

Answer: (A)

5. Solution 1Draw a line from M to T on SR so that MT is parallel to QR.Then MTRQ is a rectangle. This means that the area of 4MQRis half of the area of rectangle MTRQ.Thus, the area of MTRQ is 2× 100 = 200.Since M is the midpoint of PQ and PQRS is a square, then T isthe midpoint of SR.This means that the area of MTRQ is half of the area of PQRS.Therefore, the area of PQRS is 2× 200 = 400.

P S

RQ

M T

Solution 2Suppose that the side length of square PQRS is 2x.Since M is the midpoint of PQ, then MQ = 1

2(2x) = x.

Since PQRS is a square, then 4MQR is right-angled at Q.Therefore, the area of 4MQR is 1

2(MQ)(QR) = 1

2(x)(2x) = x2.

Since the area of 4MQR is 100, then x2 = 100, and so x = 10, since x > 0.Thus, the side length of square PQRS is 2x = 20 and so the area of square PQRS is 202 = 400.

Answer: (D)

6. Suppose that John ate x peanuts on the fourth night.Since he ate 6 more peanuts each night than on the previous night, then he ate x− 6 peanutson the third night, (x− 6)− 6 = x− 12 peanuts on the second night, and (x− 12)− 6 = x− 18peanuts on the first night.Since John ate 120 peanuts in total, then x + (x − 6) + (x − 12) + (x − 18) = 120, and so4x− 36 = 120 or 4x = 156 or x = 39.Therefore, John ate 39 peanuts on the fourth night.

Answer: (B)

7. Suppose that the side length of each of the five identical squares is x.Then PS = QR = x and PQ = SR = 5x.Since the perimeter of rectangle PQRS is 48, then 5x+ x+ 5x+ x = 48 or 12x = 48 or x = 4.Therefore, PS = QR = 4 and PQ = SR = 5 · 4 = 20, and so the area of rectangle PQRS is20 · 4 = 80.

Answer: (C)


8. Since v = 3x and x = 2, then v = 3 · 2 = 6.Therefore, (2v − 5)− (2x− 5) = (2 · 6− 5)− (2 · 2− 5) = 7− (−1) = 8.

Answer: (B)

9. Suppose that Sally’s original height was s cm.Since Sally grew 20% taller, her new height is 1.2s cm.Since Sally is now 180 cm tall, then 1.2s = 180 or s = 180

1.2= 150.

Thus, Sally grew 180− 150 = 30 cm.Since Mary grew half as many centimetres as Sally grew, then Mary grew 1

2· 30 = 15 cm.

Since Mary and Sally were originally the same height, then Mary was originally 150 cm tall,and so is now 150 + 15 = 165 cm tall.

Answer: (B)

10. Since (2a)(2b) = 64, then 2a+b = 64, using an exponent law.Since 64 = 26, then 2a+b = 26 and so a+ b = 6.Therefore, the average of a and b is 1

2(a+ b) = 3.

Answer: (D)

11. If N is divisible by both 5 and 11, then N is divisible by 5× 11 = 55.This is because 5 and 11 have no common divisor larger than 1.Therefore, we are looking for a multiple of 55 between 400 and 600 that is odd.One way to find such a multiple is to start with a known multiple of 55, such as 550.We can add or subtract 55 from this multiple and still obtain multiples of 55.Note that 550 + 55 = 605, which is too large.Now 550− 55 = 495 which is in the correct range and is odd.Since we are told that there is only such such integer, then it must be the case that N = 495.The sum of the digits of N is 4 + 9 + 5 = 18.

Answer: (E)

12. Since 4QUR and 4SUR are equilateral, then ∠QUR = ∠SUR = 60◦.Since QU = PU = TU = SU and QP = PT = TS, then 4QUP , 4PUT and 4TUS arecongruent.Thus, ∠QUP = ∠PUT = ∠TUS.The angles around point U add to 360◦.Thus, ∠SUR+ ∠QUR+ ∠QUP + ∠PUT + ∠TUS = 360◦ and so 60◦ + 60◦ + 3∠TUS = 360◦

or 3∠TUS = 240◦ or ∠TUS = 80◦.Since 4TUS is isosceles with TU = SU , then ∠UST = ∠UTS.Since the angles in 4TUS add to 180◦, then ∠TUS + ∠UST + ∠UTS = 180◦.Therefore, 80◦ + 2∠UST = 180◦ and so 2∠UST = 100◦ or ∠UST = 50◦.

Answer: (A)

13. The quilt consists of 25 identical squares.Of the 25 squares, 4 are entirely shaded, 8 are shaded with a single triangle that covers half ofthe square, and 4 are shaded with two triangles that each cover a quarter of the square.Therefore, the shading is equivalent to the area of 4 + 8× 1

2+ 4× 2× 1

4= 10 squares.

As a percentage, the shading is 1025× 100% = 40% of the total area of the quilt.

Answer: (B)


14. Solution 1Since the two terms have a common factor, then we factor and obtain (x−2)((x−4)+(x−6)) = 0.This gives (x− 2)(2x− 10) = 0.Therefore, x− 2 = 0 (which gives x = 2) or 2x− 10 = 0 (which gives x = 5).Therefore, the two roots of the equation are x = 2 and x = 5. Their product is 10.

Solution 2We expand and then simplify the left side:

(x− 4)(x− 2) + (x− 2)(x− 6) = 0

(x2 − 6x+ 8) + (x2 − 8x+ 12) = 0

2x2 − 14x+ 20 = 0

Since the product of the roots of a quadratic equation of the form ax2 + bx+ c = 0 with a 6= 0

isc

a, then the product of the roots of the equation 2x2 − 14x+ 20 = 0 is

20

2= 10.

Answer: (C)

15. Because of the way in which the oranges are stacked, each layer is a rectangle whose length is1 orange less and whose width is 1 orange less than the layer below.The bottom layer is 5 by 7 and so contains 35 oranges.The next layer is 4 by 6 and so contains 24 oranges.The next layer is 3 by 5 and so contains 15 oranges.The next layer is 2 by 4 and so contains 8 oranges.The next layer is 1 by 3 and so contains 3 oranges. This is the last layer, as it consists of asingle row of oranges.The total number of oranges in the stack is thus 35 + 24 + 15 + 8 + 3 = 85.

Answer: (D)

16. Since there are 30 people in a room and 60% of them are men, then there are 610× 30 = 18 men

in the room and 12 women.Since no men enter or leave the room, then these 18 men represent 40% of the final number inthe room.Thus, 9 men represent 20% of the the final number in the room, and so the final number ofpeople is 5× 9 = 45.Since 18 of these are men and 12 of these are the women originally in the room, then45− 18− 12 = 15 women entered the room.

Answer: (E)

17. Since 32011 = 31 · 32010 = 3 · 32010 and 32012 = 32 · 32010 = 9 · 32010, then

32011 + 32011

32010 + 32012=

3 · 32010 + 3 · 32010

32010 + 9 · 32010=

32010(3 + 3)

32010(1 + 9)=

3 + 3

1 + 9=

6

10=

3

5Answer: (A)

18. In order to find N , which is the smallest possible integer whose digits have a fixed product, wemust first find the minimum possible number of digits with this product. (This is because ifthe integer a has more digits than the integer b, then a > b.)Once we have determined the digits that form N , then the integer N itself is formed by writingthe digits in increasing order. (Given a fixed set of digits, the leading digit of N will contribute


to the largest place value, and so should be the smallest digit. The next largest place valueshould get the next smallest digit, and so on.)Note that the digits of N cannot include 0, else the product of its digits would be 0.Also, the digits of N cannot include 1, otherwise we could remove the 1s and obtain an integerwith fewer digits (thus, a smaller integer) with the same product of digits.

Since the product of the digits of N is 1728, we find the prime factorization of 1728 to help usdetermine what the digits are:

1728 = 9× 192 = 32 × 3× 64 = 33 × 26

We must try to find a combination of the smallest number of possible digits whose product is1728.Note that we cannot have 3 digits with a product of 1728 since the maximum possible productof 3 digits is 9× 9× 9 = 729.Let us suppose that we can have 4 digits with a product of 1728.In order for N to be as small as possible, its leading digit (that is, its thousands digit) must beas small as possible.From above, this digit cannot be 1.This digit also cannot be 2, since otherwise the product of the remaining 3 digits would be 864,which is larger than the product of 3 digits can be.Can the thousands digit be 3? If so, the remaining 3 digits have a product of 576.Can 3 digits have a product of 576?If one of these 3 digits were 7 or less, then the product of the 3 digits would be at most7× 9× 9 = 567, which is too small.Therefore, if we have 3 digits with a product of 576, then each digit is 8 or 9.Since the product is even, then at least one of the digits would have to be 8, leaving theremaining two digits to have a product of 576÷ 8 = 72.These two digits would then have to be 8 and 9.Thus, we can have 3 digits with a product of 576, and so we can have 4 digits with a productof 1728 with smallest digit 3.Therefore, the digits of N must be 3, 8, 8, 9. The smallest possible number formed by thesedigits is when the digits are placed in increasing order, and so N = 3889.The sum of the digits of N is 3 + 8 + 8 + 9 = 28.

Answer: (A)

19. We label the three points as O(0, 0), P (1, 4) and Q(4, 1).There are three possible locations for the fourth vertex R of the parallelogram – between Oand P (in the second quadrant), between P and Q (in the first quadrant), and between Q andO (in the fourth quadrant).In each of these cases, 4OPQ will make up half of the parallelogram, and so the area of theparallelogram is twice the area of 4OPQ.There are many ways to calculate the area of 4OPQ.We proceed by “completing the rectangle” which includes thex-axis, the y-axis, the line y = 4, and the line x = 4.We label the point (0, 4) as S, the point (4, 4) as T , and the point(4, 0) as U .(Note that rectangle OSTU is in fact a square, so we have “com-pleted the square”!)The area of 4OPQ equals the area of rectangle OSTU minus thecombined areas of 4OSP , 4PTQ, and 4QUO.

xQ (4, 1)

P (1, 4)

(0, 0)

y

O

S T

U


The area of rectangle OSTU is 4 · 4 = 16, since it is a square with side length 4.Consider 4OSP . It is right-angled at S, with OS = 4 and SP = 1.Thus, its area is 1

2(OS)(SP ) = 1

2(4)(1) = 2.

Consider 4PTQ. It is right-angled at T , with PT = TQ = 3.Thus, its area is 1

2(PT )(TQ) = 1

2(3)(3) = 9

2.

Consider 4QUO. It is right-angled at U , with OU = 4 and UQ = 1.Thus, its area is 1

2(OU)(UQ) = 1

2(4)(1) = 2.

Therefore, the area of 4OPQ is 16− 2− 92− 2 = 15

2.

Thus, the area of the parallelogram is 2 · 152

= 15.

Answer: (A)

20. In the first race, Katie ran 100 m in the same time that Sarah ran 95 m.This means that the ratio of their speeds is 100 : 95 = 20 : 19.In other words, in the time that Sarah runs 1 m, Katie runs 20

19≈ 1.053 m.

Put another way, in the time that Katie runs 1 m, Sarah runs 1920

= 0.95 m.In the second race, Katie must run 105 m and Sarah must run 100 m.If Sarah finishes first, then Katie must not have completed 105 m in the time that it takesSarah to complete 100 m.But Katie runs 1.053 m for 1 m that Sarah runs, so Katie will in fact run more than 105 m inthe time that Sarah runs 100 m.Therefore, Katie must finish first.In the time that Katie runs 105 m, Sarah will run 105× 19

20= 1995

20= 399

4= 993

4.

Thus, Sarah was 100− 9934

= 14

= 0.25 m behind.Therefore, when Katie crossed the finish line, Sarah was 0.25 m behind.

Answer: (B)

21. Since x2 = 8x+ y and y2 = x+ 8y, then x2 − y2 = (8x+ y)− (x+ 8y) = 7x− 7y.Factoring both sides, we obtain (x+ y)(x− y) = 7(x− y).Since x 6= y, then x− y 6= 0, so we can divide both sides by x− y to obtain x+ y = 7.Since x2 = 8x+ y and y2 = x+ 8y, then

x2 + y2 = (8x+ y) + (x+ 8y) = 9x+ 9y = 9(x+ y) = 9 · 7 = 63

Answer: (C)

22. From the chart, we see that QR = 25, QS = 7 and SR = 18.Since QR = QS + SR and QR is the largest of these three lengths, then S must be a point online segment QR.This gives the following configuration so far:

Q S R187

We have not yet used the fact that PQ = 25 or that PS = 24.Note that 72 + 242 = 49 + 576 = 625 = 252, so QS2 + PS2 = PQ2.Since these lengths satisfy this property, then the points P , S and Q form a triangle that isright-angled at S.This gives the following configuration so far:


Q S R187

2425

P

(We could have drawn P “above” QR.)Since ∠PSQ = 90◦, then ∠PSR = 90◦.Therefore, PR2 = PS2 + SR2 = 242 + 182 = 576 + 324 = 900.Since PR > 0, then PR =

√900 = 30.

Thus, the distance between cities P and R is 30.Answer: (A)

23. Initially, the bowl contains 320 g of white sugar and 0 g of brown sugar.Mixture Y contains (320− x) g of white sugar and x g of brown sugar.When Mixture Z (the final mixture) is formed, there is still 320 g of sugar in the bowl.Since we are told that the ratio of the mass of white sugar to the mass of brown sugar is 49 : 15,then the mass of white sugar in Mixture Z is 49

49+15· 320 = 49

64· 320 = 49 · 5 = 245 g and the

mass of brown sugar in Mixture Z is 320− 245 = 75 g.In order to determine the value of x (and hence determine the values of w and b), we need todetermine the mass of each kind of sugar in Mixture Z in terms of x.Recall that Mixture Y consists of (320− x) g of white sugar and x g of brown sugar, which arethoroughly mixed together.

Because Mixture Y is thoroughly mixed, then each gram of Mixture Y consists of320− x

320g of

white sugar andx

320g of brown sugar.

To form Mixture Z, x g of Mixture Y are removed.

This amount of Mixture Y that is removed contains x · x

320=

x2

320g of brown sugar.

Mixture Z is made by removing x g of Mixture Y (which containsx2

320g of brown sugar),then

adding x g of brown sugar.

Thus the mass of brown sugar, in g, in Mixture Z is x− x2

320+ x.

Since Mixture Z includes 75 g of brown sugar, then

2x− x2

320= 75

0 = x2 − 2(320)x+ 75(320)

0 = x2 − 640x+ 24000

0 = (x− 40)(x− 600)

Therefore, x = 40 or x = 600.Since the initial mixture consists of 320 g of sugar, then x < 320, so x = 40.This tells us that Mixture Y consists of 320 − 40 = 280 g of white sugar and 40 g of brownsugar. The ratio of these masses is 280 : 40, which equals 7 : 1 in lowest terms. Thus, w = 7and b = 1.Therefore, x+ w + b = 40 + 7 + 1 = 48.

Answer: (A)


24. We use without proof the fact that if a circle with cen-tre O and radius r touches (that is, is tangent to)line segments AB, BC and CD at X, Y and Z, re-spectively, then ∠OBX = ∠OBY = 1

2(∠ABC) and

∠OCY = ∠OCZ = 12(∠BCD).

Suppose that ∠ABC = θ and ∠BCD = α.

Since OY is perpendicular to BC, then tan(∠OBY ) =OY

BY

and tan(∠OCY ) =OY

Y C.

A

B C

D

X

Y

ZO

Thus, BY =OY

tan(∠OBY )=

r

tan(θ/2)and Y C =

OY

tan(∠OCY )=

r

tan(α/2).

Since BC = BY + Y C, then

BC =r

tan(θ/2)+

r

tan(α/2)

BC = r

(1

tan(θ/2)+

1

tan(α/2)

)BC = r

(tan(θ/2) + tan(α/2)

tan(θ/2) tan(α/2)

)r =

BC tan(θ/2) tan(α/2)

tan(θ/2) + tan(α/2)

Consider now the given quadrilateral QRST . We know thatTQ = 3. Since 4PQR is equilateral, then QR = PQ and soQR = PT + TQ = 1 + 3 = 4.Since PR = QR = 4 and S is the midpoint of PR, then RS = 2.Since4PST has PT = 1 and PS = 1

2PR = 2 and ∠SPT = 60◦,

then it is a 30◦-60◦-90◦ triangle, so we have ST =√

3.Also, we have ∠PTS = 90◦ and ∠PST = 30◦, which give∠STQ = 90◦ and ∠RST = 150◦.To summarize, quadrilateral QRST has QR = 4, RS = 2,ST =

√3, TQ = 3, ∠TQR = 60◦, ∠QRS = 60◦, ∠RST = 150◦,

and ∠STQ = 90◦.

P

QR

S

T2

2

1

3

460 60

90

150

Next, we determine the radius of a circle tangent to each set of three consecutive sides, ignoringthe fact that the fourth side might restrict the size of the circle. After determining these radii,we examine how the fourth side comes into play.Suppose that we have a circle tangent to line segments TQ, QR and RS.By the formula above, the radius of this circle would equal

4 tan(60◦/2) tan(60◦/2)

tan(60◦/2) + tan(60◦/2)=

4 tan(30◦) tan(30◦)

tan(30◦) + tan(30◦)≈ 1.1547

Suppose that we have a circle tangent to line segments QR, RS and ST .By the formula above, the radius of this circle would equal

2 tan(60◦/2) tan(150◦/2)

tan(60◦/2) + tan(150◦/2)=

2 tan(30◦) tan(75◦)

tan(30◦) + tan(75◦)= 1


Suppose that we have a circle tangent to line segments RS, ST and TQ.By the formula above, the radius of this circle would equal

√3 tan(150◦/2) tan(90◦/2)

tan(150◦/2) + tan(90◦/2)=

√3 tan(75◦) tan(45◦)

tan(75◦) + tan(45◦)≈ 1.3660

Suppose that we have a circle tangent to line segments ST , TQ and QR.By the formula above, the radius of this circle would equal

3 tan(90◦/2) tan(60◦/2)

tan(90◦/2) + tan(60◦/2)=

3 tan(45◦) tan(30◦)

tan(45◦) + tan(30◦)≈ 1.0981

We need to determine the radius of the largest circle that can be drawn inside quadrilateralQRST .The largest such circle will be touching at least two adjacent sides of QRST . Why is this? If acircle were touching zero sides or one side of QRST , it could be slid until it was touching twoconsecutive sides and then expanded a little bit and so cannot be the largest such circle. If itwere touching two opposite sides of QRST , then it could be slid to touch a side adjacent toone of these two sides, perhaps losing contact with one of the two opposite sides.Consider now a circle touching two adjacent sides of QRST . Such a circle can be expandedwhile maintaining contact to these two sides until it touches a third side. Once it touches thethird side, it can’t be expanded any further because its radius is fixed by the calculation thatwe did at the beginning of the solution.Therefore, the largest circle will be touching three of the sides of QRST .

In order to complete the solution, we need to determine which of the circles that touch threesides actually lie completely inside QRST .We do this by examining each of the four pairs of consecutive sides, and determining what thelargest circle is that can be drawn touching these sides.

– We consider a circle tangent to ST and TQ. We expand the circle, keeping it tangent toST and TQ, until it touches either QR or RS. From the previous calculations, the circlethat also touches QR has radius about 1.0981, and the circle that also touches RS hasradius about 1.3660. Thus, the circle will first touch QR. In this case, the largest circlethat is completely inside the quadrilateral has radius 1.0981.

– We consider a circle tangent to TQ and QR. We expand the circle, keeping it tangent toTQ and QR, until it touches either ST or RS. From the previous calculations, the circlethat also touches ST has radius about 1.0981, and the circle that also touches RS hasradius about 1.1547. Thus, the circle will first touch ST . In this case, the largest circlethat is completely inside the quadrilateral has radius 1.0981.

– We consider a circle tangent to QR and RS. We expand the circle, keeping it tangentto QR and RS, until it touches either ST or TQ. From the previous calculations, thecircle that also touches ST has radius 1, and the circle that also touches TQ has radiusabout 1.1547. Thus, the circle will first touch ST . In this case, the largest circle that iscompletely inside the quadrilateral has radius 1.

– We consider a circle tangent to RS and ST . We expand the circle, keeping it tangent toRS and ST , until it touches either QR or TQ. From the previous calculations, the circlethat also touches QR has radius about 1, and the circle that also touches TQ has radiusabout 1.3660. Thus, the circle will first touch QR. In this case, the largest circle that iscompletely inside the quadrilateral has radius 1.


Finally, comparing the four cases, we see that the largest circle that we can obtain has radiusabout 1.0981, which is closest to 1.10.

Answer: (B)

25. Let N be an arbitrary positive integer with the desired properties.Let S(N) represent the sum of the digits of N and let S(2N) represent the sum of the digitsof 2N . In the table below, we make a claim about how each digit of N contributes to S(2N).We use the data in the table to answer the question, following which we justify the data in thetable:

Digit in N 2×Digit Contribution to S(2N)3 6 64 8 85 10 1 + 0 = 16 12 1 + 2 = 3

Suppose that the digits of N include w 3s, x 4s, y 5s, and z 6s. Note that w, x, y, z ≥ 1.

Step 1: Using information about S(N) and S(2N)

Since S(N) = 900, then 3w + 4x+ 5y + 6z = 900.Since each 3 in N contributes 6 to S(2N), each 4 in N contributes 8 to S(2N), each 5 in Ncontributes 1 to S(2N), and each 6 in N contributes 3 to S(2N), then S(2N) = 900 tells usthat 6w + 8x+ y + 3z = 900.

Step 2: Understanding which values of N will be largest possible and smallest possible

The largest possible value of N will be the integer N+ that satisfies the given properties, hasthe largest number of digits (that is, the largest value of w + x+ y + z), has the largest actualdigits given this fixed number of digits, and has its digits in decreasing order from left to right(since the larger digits will correspond to the largest place values).The smallest possible value of N will be the integer N− that satisfies the given properties, hasthe smallest number of digits (that is, the smallest value of w + x + y + z), has the smallestactual digits given this fixed number of digits, and has its digits in increasing order from leftto right.Since we want to determine the number of digits in the product N+N−, we mostly care onlyabout the number of digits in N+ and N− (that is, the maximum and minimum values ofw + x+ y + z), and perhaps as well about the leading digits of each.

Step 3: Simplifying equationsWe have that w, x, y, z are positive integers that satisfy both 3w + 4x + 5y + 6z = 900and 6w + 8x+ y + 3z = 900.Multiplying the first of these equations by 2, we obtain 6w + 8x+ 10y + 12z = 1800.When we subtract the second equation from this, we obtain 9y + 9z = 900, or y + z = 100.Since y + z = 100, then 3w + 4x + 5y + 6z = 900 becomes 3w + 4x + 5(y + z) + z = 900 or3w + 4x+ 500 + z = 900 or 3w + 4x+ z = 400.Also since y + z = 100, we obtain w + x+ y + z = w + x+ 100, so to minimize and maximizew + x+ y + z, we have to minimize and maximize w + x.


Step 4: Restating part of goalWe want to find the maximum and minimum possible values of w + x + y + z subject tothe conditions that w, x, y, z are positive integers that satisfy 3w + 4x + 5y + 6z = 900 and6w + 8x+ y + 3z = 900.From Step 3, these equations are true if and only if y + z = 100 and 3w + 4x+ z = 400. (Thisis because we can use each pair of equations to obtain the other pair.)Therefore, we want to find the maximum and minimum possible values of w+x+ y+ z subjectto the conditions y + z = 100 and 3w + 4x+ z = 400.Since y + z is fixed, this is the same as finding the maximum and minimum possible values ofw + x subject to the conditions y + z = 100 and 3w + 4x+ z = 400.

Step 5: Determining information about N+

We determine the maximum possible value of w + x.We rewrite 3w + 4x+ z = 400 as 3(w + x) = 400− x− z.To make w + x as large as possible, we want the right side of this equation to be as large aspossible, and so we want x and z to be as small as possible.Note that x ≥ 1 and z ≥ 1 and so 400− x− z ≤ 398.Also, since the left side of 3(w+ x) = 400− x− z is divisible by 3, then the right side must bedivisible by 3.The largest multiple of 3 less than or equal to 398 is 396.Therefore, 3(w + x) ≤ 396 and so w + x ≤ 132.Thus, the maximum possible value of w + x+ y + z is 132 + 100 = 232.To achieve this maximum, we need 400−x− z = 396 (that is, x+ z = 4). The values w = 129,x = 3, y = 99, z = 1 achieve this maximum (and satisfy y + z = 100 and 3w + 4x+ z = 400).Therefore, N+, the largest possible value of N , consists of 232 digits (all 3s, 4s, 5s and 6sincluding at least one of each) arranged in descending order.Thus, N+ satisfies 6× 10231 < N+ < 7× 10231.(As it turns out, we will not actually have to determine the actual digits of N+.)

Step 6: Determining information about N−

We determine the minimum possible value of w + x.We rewrite 3w + 4x+ z = 400 as 4(w + x) = 400 + w − z.To make w + x as small as possible, we want the right side of this equation to be as small aspossible, and so we want w to be as small as possible and z to be as large as possible.Now w ≥ 1 and since y + z = 100 and y ≥ 1, then z ≤ 99.Thus, 400 + w − z ≥ 302.Since the left side of the equation 4(w + x) = 400 + w − z is divisible by 4, then the right sidemust be divisible by 4.The smallest multiple of 4 greater than or equal to 302 is 304.Therefore, 4(w + x) ≥ 304 and so w + x ≥ 76.Thus, the minimum possible value of w + x+ y + z is 76 + 100 = 176.To achieve this minimum, we need 400 +w− z = 304 (that is, z −w = 96). The values w = 3,x = 73, y = 1, z = 99 achieve this minimum (and satisfy y + z = 100 and 3w + 4x+ z = 400).Therefore, N− consists of 176 digits (all 3s, 4s, 5s and 6s including at least one of each) arrangedin ascending order.Thus, N− satisfies 3 × 10175 < N− < 4 × 10175, since this value of N begins with a 3 and has176 digits.(Again, we will not actually have to determine the actual digits of N−.)


Step 7: Determining the number of digits in N− ·N+

Finally, since 6× 10231 < N+ < 7× 10231 and 3× 10175 < N− < 4× 10175, then

18× 10406 = (3× 10175) · (6× 10231) < N− ·N+ < (4× 10175) · (7× 10231) = 28× 10406

Therefore, N− ·N+ has 408 digits.

Justification of data in tableWe must still justify the data in the table above.Suppose that N ends with the digits abcd. That is, N = · · · dcba.Then we can write N = · · ·+ 1000d+ 100c+ 10b+ a.Then 2N = · · · + 1000(2d) + 100(2c) + 10(2b) + (2a). The difficulty in determining the digitsof 2N is that each of 2a, 2b, 2c and 2d may not be a single digit.We use the notation u(2a) and t(2a) to represent the units digit and tens digit of 2a, respectively.Note that u(2a) is one of 0, 2, 4, 6, or 8, and t(2a) is 0 or 1.We define u(2b), t(2b), u(2c), t(2c), u(2d), t(2d) similarly.Note that 2a = 10 · t(2a) + u(2a) and 2b = 10 · t(2b) + u(2b) and 2c = 10 · t(2c) + u(2c) and2d = 10 · t(2d) + u(2d).Thus,

2N = · · ·+ 1000(10 · t(2d) + u(2d)) + 100(10 · t(2c) + u(2c))

+10(10 · t(2b) + u(2b)) + (10 · t(2a) + u(2a))

= · · ·+ 1000(u(2d) + t(2c)) + 100(u(2c) + t(2b)) + 10(u(2b) + t(2a)) + u(2a)

Since u(2a), u(2b), u(2c), u(2d) ≤ 8 and t(2a), t(2b), t(2c), t(2d) ≤ 1, then each of u(2d) + t(2c)and u(2c) + t(2b) and u(2b) + t(2a) and u(2a) is a single digit, so these are the thousands,hundreds, tens and units digits, respectively, of 2N .Thus, the sum of the digits of 2N is

u(2a) + (u(2b) + t(2a)) + (u(2c) + t(2b)) + (u(2d) + t(2c)) + · · · =

(t(2a) + u(2a)) + (t(2b) + u(2b)) + (t(2c) + u(2c)) + · · ·

The above argument extends to the left for the remaining digits of N .In other words, if m is a digit in N , then its contribution to the sum of the digits of 2N is thesum of the tens and units digits of 2m.Therefore, the digits of N contribute to the sum of the digits of 2N as outlined in the tableabove.

Answer: (A)

2012 fermat contest - · pdf file2012 fermat contest solutions page 2 1. since 60 8 = 60 8 =...

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